Think geometrically. Do you know about Euler's formula? It states that
$$e^{i\theta} = \cos\theta + i \sin\theta$$
By comparing, you can see that your number is $w = e^{2\pi i/5}$, which geometrically corresponds to the point in the argand diagram a distance 1 from the origin, with argument $2\pi/5$, which is 1/5 of the way round a circle.
Squaring to get $w^2$ leaves the modulus unchanged and doubles the argument, so the point $w^2$ is 2/5 of the way round the circle, and $w^3$ and $w^4$ are 3/5 and 4/5 of the way round the circle.
Try plotting the points $1, w, w^2, w^3, w^4$ on the argand diagram and see what you notice about them. Now can you start adding them up?
Let $z=a+ib$.
First, the two absolute values are
$$|z-1|=\sqrt{(a-1)^2+b^2} \text{ and } |z-i|=\sqrt{a^2+(b-1)^2}.$$
We have the following inequality then
$$a^2-2a+1+b^2\le a^2+b^2-2b+1$$
or
$$-2a\le -2b$$
or $$a\ge b.$$
That is, the first inequality holds for those complex numbers whose real part is greater or equal than their imaginary par.
As far as the second inequality we have
$$|z-(2+2i)|=\sqrt{(a-2)^2+(b-2)^2}.$$
So we have the following inequality
$$(a-2)^2+(b-2)^2\le 1$$
which holds inside the unit circle centered at $2+i2$.
The following two figure depicts the two regions
The third figure depicts the common region in red. The red line is tangent to the circle.
The angle between the tangent line and the Real axis is the smallest argument belonging to complex numbers satisfying both of the inequalities.
Further edit:
The length of the segment (the absolute value of the complex number sought) joining the origin and the tangent point is $\sqrt7$ as explained by the figure below
Best Answer
If you translate the inequalities to what they say about $x$ and $y$, you can use Wolfram Alpha. I don't know if you can use Wolfram Alpha directly: if you give it an inequality involving complex numbers, it just says "Inequalities are not well-defined in the complex plane".
In Maple you could use something like this: